Question

if a-1/a=15/4 , find the value of a+1/a please give answers with steps

Answer 1

☆Given:-

•Value of (a-1/a)=15/4

☆To find:-

•Value of (a+1/a)

☆Solution:-

Here,we shall apply the formula:-

=>(x+y)²=(x-y)²+4xy

=>(a+1/a)²=(a-1/a)²+4×a×1/a

=>(a+1/a)²=(15/4)²+4

=>(a+1/a)²=225/16+4

=>(a+1/a)²=(225+64)/16

=>(a+1/a)²=289/16

$= > (a + \frac{1}{a} ) = \sqrt{ \frac{289}{16} }$

=>(a+1/a)=17/4

Thus,value of (a+1/a) is 17/4

☆Some identities, used to solve this type of problems:-

•(a+b)²=a²+b²+2ab

•(a-b)²=a²+b²-2ab

•a²-b²=(a-b)(a+b)

•(a+b+c)²=a²+b²+c²+2(ab+bc+ac)

•(a+b)³=a³+b³+3ab(a+b)

•(a-b)³=a³-b³-3ab(a-b)

•a³+b³=(a+b)(a²+b²-ab)

•a³-b³=(a-b)(a²+b²+ab)

Answer 2

Answer:

Given:

• $a - \dfrac{1}{a}= \dfrac{15}{4}$

To find:

• $a + \dfrac{1}{a}$

$\sf{\underline{Solution:-}}$

In order to solve this , we will first square both sides and then make positive square and hence find the value .

☘Identity used :

$\boxed{(a-b)^2 = a^2 + b^2 + 2ab}$

✦Now squaring both sides:

➫$\left( a - \dfrac{1}{a} \right)^2 = \left(\dfrac{15}{4}\right)^2$

➫$a^2 + \dfrac{1}{a^2} - 2 = \dfrac{225}{16}$

✦Adding 4 on both sides:

➫$a^2 + \dfrac{1}{a^2} - 2 + 4 = \dfrac{225}{16} + 4$

➫$a^2 + \dfrac{1}{a^2} + 2 = \dfrac{225 + 64}{16}$

➫$a^2 + \dfrac{1}{a^2} + 2 = \dfrac{289}{16}$

☘Identity used:

$\boxed{(a+b)^2 = a^2 + b^2 + 2ab}$

✦Taking square root both sides:

➫$\left(a + \dfrac{1}{a} \right)= \sqrt{\dfrac{289}{16}}$

➫$a + \dfrac{1}{a} = \dfrac{17}{4}$